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Let $\pi$ be the joint law of $(X, Y)$ with marginal distributions $\mu$ and $\nu$. We assume that we have: for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$

$$ \nu\left(\{y \in \mathbb{R} : \pi_{Y=y}(A) = 1\}\right) > 0. $$ where $\pi_{Y=y}$ is the conditional law of $X$ given $Y=y$. I am asking if necessarily:

  • $\nu\left(\{y \in \mathbb{R} : \pi_{Y=y} \text{ is a Dirac measure}\}\right) > 0$,
  • More precisely, if for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$:

$$ \nu\left(\{y \in \mathbb{R} : \pi_{Y=y}(A) = 1 \text{ and } \pi_{Y=y} \text{ is a Dirac measure}\}\right) > 0. $$

I am unable to prove this or find any counterexample. Thank you very much for your help.

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  • $\begingroup$ If $Y$ is deterministic, is this known to be true/false? @thibualt jeannin $\endgroup$
    – Nate River
    Commented Oct 13 at 15:10
  • $\begingroup$ It seems like the question is entirely determined by the case of deterministic $Y$, since we can just use the regular conditional probability given $Y$ to make $Y$ deterministic? $\endgroup$
    – Nate River
    Commented Oct 13 at 15:14
  • $\begingroup$ Indeed, would that give you what you wanted? @thibault jeannin $\endgroup$
    – Nate River
    Commented Oct 13 at 18:55
  • $\begingroup$ Since a deterministic function is a Dirac delta, so then $\nu(\{y \, | \, \mu_{X|Y} \text{ is a Dirac measure}\})$ is equal to $1$. $\endgroup$
    – Nate River
    Commented Oct 13 at 18:56
  • $\begingroup$ @NateRiver thanks you very much. I am a bit confused, how should I generalize from the case Y deterministic? $\endgroup$
    – thibault jeannin
    Commented Oct 13 at 20:13

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